Solution for a second order ODE with initial conditions
Solution 1:
$$\dot{\theta}(0)=D\sin(E)+D\frac{\sqrt{4K-C^2}}{2}\cos(E)=0$$
Divide by $D \sin E$: $$\frac{\sqrt{4K-C^2}}{2} \dfrac {\cos E}{\sin E}=-1$$ $$\tan E= -\frac{\sqrt{4K-C^2}}{2}$$
$$\dot{\theta}(0)=D\sin(E)+D\frac{\sqrt{4K-C^2}}{2}\cos(E)=0$$
Divide by $D \sin E$: $$\frac{\sqrt{4K-C^2}}{2} \dfrac {\cos E}{\sin E}=-1$$ $$\tan E= -\frac{\sqrt{4K-C^2}}{2}$$